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Question-69416




Question Number 69416 by ahmadshah last updated on 23/Sep/19
Commented by mr W last updated on 23/Sep/19
not defined for real numbers!  domain of function x^(√2)  is x≥0.
$${not}\:{defined}\:{for}\:{real}\:{numbers}! \\ $$$${domain}\:{of}\:{function}\:{x}^{\sqrt{\mathrm{2}}} \:{is}\:{x}\geqslant\mathrm{0}. \\ $$
Answered by mr W last updated on 23/Sep/19
−1=−1+0i=cos π+i sin π=e^(iπ)   (−1)^(√2) =e^(i(√2)π) =cos ((√2)π)+i sin ((√2)π)  =−0.266255−0.963903 i
$$−\mathrm{1}=−\mathrm{1}+\mathrm{0}{i}=\mathrm{cos}\:\pi+{i}\:\mathrm{sin}\:\pi={e}^{{i}\pi} \\ $$$$\left(−\mathrm{1}\right)^{\sqrt{\mathrm{2}}} ={e}^{{i}\sqrt{\mathrm{2}}\pi} =\mathrm{cos}\:\left(\sqrt{\mathrm{2}}\pi\right)+{i}\:\mathrm{sin}\:\left(\sqrt{\mathrm{2}}\pi\right) \\ $$$$=−\mathrm{0}.\mathrm{266255}−\mathrm{0}.\mathrm{963903}\:{i} \\ $$

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